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-3=cot^2(θ)+3csc(θ)

解答

- 3 = cot^{2} (θ) + 3 csc (θ)

解答

θ = \frac{3 π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

+1

度数

θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

- 3 = cot^{2} (θ) + 3 csc (θ)

交换两边

cot^{2} (θ) + 3 csc (θ) = - 3

两边减去

- 3

cot^{2} (θ) + 3 csc (θ) + 3 = 0

使用三角恒等式改写

3 + cot^{2} (θ) + 3 csc (θ)

使用毕达哥拉斯恒等式:

1 + cot^{2} (x) = csc^{2} (x)

cot^{2} (x) = csc^{2} (x) - 1

= 3 + csc^{2} (θ) - 1 + 3 csc (θ)

化简

3 + csc^{2} (θ) - 1 + 3 csc (θ) : csc^{2} (θ) + 3 csc (θ) + 2

3 + csc^{2} (θ) - 1 + 3 csc (θ)

对同类项分组

= csc^{2} (θ) + 3 csc (θ) + 3 - 1

数字相加/相减:

3 - 1 = 2

= csc^{2} (θ) + 3 csc (θ) + 2

= csc^{2} (θ) + 3 csc (θ) + 2

2 + csc^{2} (θ) + 3 csc (θ) = 0

用替代法求解

2 + csc^{2} (θ) + 3 csc (θ) = 0

令

: csc (θ) = u

2 + u^{2} + 3 u = 0

2 + u^{2} + 3 u = 0 : u = - 1, u = - 2

2 + u^{2} + 3 u = 0

改写成标准形式

a x^{2} + b x + c = 0

u^{2} + 3 u + 2 = 0

使用求根公式求解

u^{2} + 3 u + 2 = 0

二次方程求根公式:

若

a = 1, b = 3, c = 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot 2 = 1

3^{2} - 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 3 + 1}{2 \cdot 1}, u_{2} = \frac{- 3 - 1}{2 \cdot 1}

u = \frac{- 3 + 1}{2 \cdot 1} : - 1

\frac{- 3 + 1}{2 \cdot 1}

数字相加/相减:

- 3 + 1 = - 2

= \frac{- 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- 3 - 1}{2 \cdot 1} : - 2

\frac{- 3 - 1}{2 \cdot 1}

数字相减:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 4}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

二次方程组的解是:

u = - 1, u = - 2

u = csc (θ)

代回

csc (θ) = - 1, csc (θ) = - 2

csc (θ) = - 1, csc (θ) = - 2

csc (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

csc (θ) = - 1

csc (θ) = - 1

的通解

csc (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

csc (θ) = - 2 : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

csc (θ) = - 2

csc (θ) = - 2

的通解

csc (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

合并所有解

θ = \frac{3 π}{2} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

流行的例子

arccos(e^{-x})= pi/3

arccos (e^{- x}) = \frac{π}{3}

cos(θ)=-(11)/(sqrt(170))

cos (θ) = - \frac{11}{170}

2cos^2(θ)-7cos(θ)+3=0,0<= θ<= 2pi

2 cos^{2} (θ) - 7 cos (θ) + 3 = 0, 0 \leq θ \leq 2 π

4sin(x)cos(x)+2sin(x)-2cos(x)=1

4 sin (x) cos (x) + 2 sin (x) - 2 cos (x) = 1

sin(x)cos^2(x)=sin(x)

sin (x) cos^{2} (x) = sin (x)